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## Algebra 2

### Course: Algebra 2>Unit 11

Lesson 8: Graphing sinusoidal functions

# Sinusoidal function from graph

Sal finds the equation of a sinusoidal function from its graph where the minimum point (-2,-5) and the maximum point (2,1) are highlighted. Created by Sal Khan.

## Want to join the conversation?

• Can't this graph also be represented by a cosine function with a phase shift? •   Yes, any sine function can be represented by a cosine function with a shift of pi/2. And any cos function can be represented by a sin function with a shift of pi/2.
• Hi!
I don't know if it is my stupidity or not, but to me it seems like the "Graph sinusoidal functions: phase shift" exercise block in "Graphing sinusoidal functions" has exercises that are not covered by that point, only in this video, which is next in line. Can anyone else confirm or dismiss this? •  I agree, fairly often I find that they threw an exercise in covering a topic we haven't learned yet, but is related. In other words skip it and the video, may or may not come later. Hope that's what you were looking for.
• at , Sal immediately writes the amplitude as 3. How does he know that it's positive three that not negative three? •  Amplitudes are described by their absolute value. An amplitude of -3 or an amplitude of +3 will only flip the sinusoidal function over the x-axis. The actual height of the graph is unchanged.
• There is no video for constructing a sinusoidal function and the hints in the practice are not very helpful. I might be missing something. Exactly how can the midline be inferred from two points? What is an extremum? • how do u calculate the phase shift using angles in sine and cosine • Is the period and the frequency the same thing? If not, what is the difference between them? • Period and frequency are reciprocals of each other in Physics, i.e. P = 1/f and f = 1/P. When discussing the graphs of trig functions, the Period is the length of a cycle. The term "frequency" is not formally defined. For example, sin(x) has a period of 2pi, since sin(x) = sin(x + 2pi) and it is the smallest angle for which that is true. (Adding 2pi to an angle is equivalent to one full revolution around a circle.)

To find the period of sin(bx), calculate P = 2*pi/b. For example, sin(3x) has a period of 2pi/3. Note the inverse relationship between P and "b", just as between P and "f". You could THINK of "b" as being the frequency, but it isn't formally defined.
• I am trying to do the phase shift assignment for my class but i'm really not understanding how to find the period. I can get the mid-line and amp easily enough but i get confused on the sin or cosine part and the period • In this answer, l am assuming the function involves a sine or cosine function.

If you are given the graph, then the period is the horizontal distance from one minimum to the next minimum (or from one maximum to the next maximum).

If you are instead given the equation, then the period is 2pi divided by the absolute value of the coefficient of x inside the sine or cosine function. This follows from the concept that increasing any angle by 2pi radians (a full revolution) has no effect on the angle’s sine and cosine.
• Can someone please explain y=Asin(Bx+C)+D form to me? the problem set for this gives us 2 points (may even give the midline/max/min points) and tells us to give the equation, this sample problem from the video gives that graph which makes it so much easier. Is there an easy way to do these type of problems (find the equation given 2 points)? Thanks in advance • f(x) = Asin(bx+c)+D — breaking it down:
A = Amplitude
B (assuming it's 1) = Period
C = Horizontal Shift
D = Vertical Shift/midline

Typically in "find the equation given 2 points," your first goal is finding the period. An useful graph @ the hints in the practice problems tremendously aided me, so I recommend using that. For simple:

If given mid --> minimum/maximum, take the x distance and multiply by four. Then take 2pi/(that #) to find B
If given min --> minimum, take the x distance and multiply by two. Then take 2pi/(tht #) to find B.

Finding A, if given mid --> mid/max, take the y distance
if given min/max --> max/min, take the y distance and divide by 2.

Finding D, find the average of the y-values of the coordinates given as max/min. Midline have might been already given.

I also recommend looking in the comments for each video, as many questions such as this has already been answered and could be more in depth than mine. Quite honestly this is objectively one of the more harder subjects in alg ii.
• Have been struggling to get the horizontal phase shift because still in April 2023 there have been no lessons/instructions on how to handle (bx + c) instead of just (kx). It is only found in the explanations to the quizzes. • When you have sin(bx+c), you're doing two things:

1. You're magnifying the argument by a factor of b and hence, you're shrinking the "width" of the function (making it more congested)

2. You're shifting the argument by c units to the left (assuming c > 0). As to why the shift is to the left, read on:

Let's take a simpler function f(x) = 2x + 2. Now, this crosses the x axis at x = -1 (or, x = -1 is the value of x which makes the function 0). Now, if I make a new function as f(x+1), I get 2(x+1) + 2. Now, the value of x at which the function becomes 0 has changed (it was initially -1. Now it's -2). To compensate for this change, the function shifts by 1 unit to the left and now crosses the x axis at -2.
• Regarding the practice section: "Graph sinusoidal functions: phase shift".
I think that this section needs more explanation, especially about (bx+c) conversion. We haven't learned how to interpret (pi*x - 3*pi) of f(x) = 2sin(pi*x - 3*pi) -4 on the graph, though I understand the horizontal shift (x-3). I understand the (bx + c) part, but can't understand when it turns into (b*pi*x + c*pi). • I find it super weird that it is using (bx+c).

When doing transformations on ALL graphs it is most useful if you have it in the form a * f(b(x-c)) + d so (bx+c) should be b(x+c/b). This is because in (bx+c) the graph is being shifted to the left by c/b units.

I'm not sure where you are getting (b*pi*x + c*pi) if you fill in the parts of a * f(b(x-c)) + d and then distribute the b into the inner part of the function you should get your answer.

a is the amplitude of course

b gets you the period, though you need to invert it. So if it goes from 2pi and doubles to 4pi you DIVIDE by 2 so b would be 1/2. usually b is the horizontal stretch/ shrink, but since sine and cosine graphs are measured in periods it's a bit easier to describe how they effect those. Still, a b between 0 and 1 is a stretch and b greater than 1 is a shrink. If it were negative then you'd have it flipped across the y axis, which you usually won't have.

c is the horizontal shift (or phase shift) but the sign is flipped. so +3 means 3 to the left and vice versa

d is the vertical shift, another easy one because plus means up and minus means down. Probably best to focus on the midline here, because it will be unaffected by the amplitude change

Once you have all that filled in the inner part goes from b(x - c) to (bx - bc)

Let me know if this doesn't help